Chapter 1 First-Order Differential Equations Shurong Sun University of Jinan Semester 1, 2010-2011.

Chapter 1Chapter 1 First-Order Differential Equations First-Order Differential Equations

Shurong SunUniversity of Jinan

Semester 1, 2010-2011

1.2

2 5 4 0x x for the unknown variable x.

In this course one of our tasks will be to solve differential equations such as 2 0y y y

for the unknown function y=y(t).

Analogous to a course in algebra and trigonometry, where a good amount of time is spent solving equations such as

1.3

1.1 Separable Differential 1.1 Separable Differential EquationsEquations

• Many first order differential equations can be written as:

dxxfdyyg

xfdxdyygxfyyg

)()(

)()( or )()( '

cdxxfdyyg )()(

•We can solve these equations by integrating:

1.4

Separable Differential EquationSeparable Differential Equation

( ) ( )dyf x g y

dx

Set 1( )( )

h yg y

and multiply by dx to obtain

( ) ( ) .h y dy f x dx

( ) ( ) .h y dy f x dx c Integrating, we have

A first order differential equation of the formis called a separable differential equation.

( ) 0.g y I.

1.5

II. 0( ) 0,g then 0( )y x

is one solution to the equation ( ) ( ).dyf x g y

dx

1.6

Example Solve the equation .dy xdx y

Solution: Separate variables and rewrite the equation in the form

.ydy xdx

Integrating, we have ydy xdx or2 2

, ( 0).2 2y x

C C

Solving for y , we obtain the solution in explicit form as 2

1 .y C x

Separable Equation ExampleSeparable Equation Example

This is a separable equation.This is a separable equation.

1.7

Separable Equation ExampleSeparable Equation Example• Solve the differential equation:

' 21 y y

21dy

dxy

arctan tan( )y x c or y x c

21dyy

dx

Solution: We first rewrite the DE in fractional form

Separate variables and rewrite the equation in the form

Integrating, we get

1.8

Example Solve the IVP 1 , ( 1) 0.3

dy yy

dx x

Solution: Separate variables and rewrite the equation

in the form .1 3

dy dxy x

Integrating, we have 1 1

1 3dy dx

y x

or

ln | 1 | ln | 3 | .y x C

Solving for y , we obtain the solution in explicit form as

1 11 ( 3) 1 ( 3), 0.C Cy e x or y C x where C e

1.9

Note that y=1 is also a solution to the equation.

So the general solution to the equation is

1 ( 3),y K x where K is an arbitrary constant.

Applying the initial condition directly, we have 0 1 2K or 1 .

2K

Thus 1 11 ( 3) ( 1).2 2

y x x

1.10

Example Solve the equation56 2 1 .

cos y

dy x xdx y e

Solution: Separate variables and rewrite the equation

in the form

Integrating, we have

or

5(cos ) (6 2 1) .yy e dy x x dx

5(cos ) (6 2 1) ,yy e dy x x dx 6 2sin .yy e x x C

1.11

Example- Newton ’ s Law of CoolingExample- Newton ’ s Law of Cooling

A copper sphere is heated to o100 .cthen placed in water that is maintained at

At time t=0 it iso30 .c

Derive an equation for the temperature T of the ball as a function of time t.

After 3 minutes the sphere ’ s temperature is o70 .c

•Newton ’ s law of cooling implies:

)30( TkdtdT

1.12

Example- Newton ’ s Law of Example- Newton ’ s Law of CoolingCooling

• Step 1 - Separate variables

dtkT

dT

30

30)(

|30|ln

ktcetT

cktT

• Step 2 - Integrate

1.13

Example- Newton ’ s Law of Example- Newton ’ s Law of CoolingCooling

• Step 3 - Determine a particular solution

3070)(

10030)0(

kt

kt

etT

ceT

1865.070

3070ln31

703070)3( 3

k

eT k

•Step 4 - Determine k

1.15

1.2 Reduction to Separable Form1.2 Reduction to Separable Form

• Sometimes first order differential equations can be made separable by a simple change of variable. Consider equations of the form:

( , )dyf x y

dx

dy yg

dx x

(1) Homogeneous Equation

is homogeneous iff ( , ) ( , ).f tx ty f x y

Remark:

1.16

• Writing the differential equation in terms of u gives:

xyu

dy duy xu u x

dx dx

•This suggests we set:

( ) ( )dy dug u u x g u

dx dx

( )dux g u u

dx

1.17

• Separating variables gives:

xdx

uugdu

)(

( )dux g u u

dx

1.18

Example: Solve 2 2 2( ) 0.xy y x dx x dy

Solution: We express (6) in the derivative form

(6)

2 22

2 ( ) 1,dy xy y x y ydx x x x

then we see that the equation (6) is homogeneous.

Now let yu

x and recall that .dy du

u xdx dx

With these substitution, equation (6) becomes 2 1.du

x udx

1.19

The above equation is separable, and , on separating the

variables and integrating, we obtain

2

1 1 , arctan ln | | .1

dv dx u x Cu x

Hence, tan(ln | | ).u x C

Finally, we substitute for and solve for

to get tan(ln | | ).

yu y

xy x x C

Also note that x=0 is a solution.

1.20

(2) Equations of the Form ( )dyG ax by

dx

Set .z ax by Then the equation is transformedinto a separable one.

Example: Solve 11 ( 2) .dyy x x y

dx

Solution: .z x y Set

Then 1 ,dz dydx dx and so 1 .dy dz

dx dx

(8)

1.21

Substituting into (8) yields 1

21

1 1 ( 2) ,

( 2) 12 ( 2) .2

dzz z or

dxdz z

z zdx z

Solving this separable equation, we obtain

2

2 ,( 2) 1

zdz dx

z

21 ln | ( 2) 1 | ,

2z x C

from which it follows that 2 2( 2) 1.xz Ce

Finally, replacing z by x-y yields2 2( 2) 1.xx y Ce

1.22

1 1 1

2 2 2

a x b y cdy fdx a x b y c

where the ,i i ia s b s and c s are constants.

I. 1 1

2 2

0.a ba b

In this case, the equations can be reduced tothe form ( )dy

G ax bydx

(3)

1.23

1 1

2 2

0.a ba b

II.

Then the system of equations 1 1 1

2 2 2

00

a x b y ca x b y c

has a unique solution 0 0( , ).x y

The above DE can be written in the form

1 0 1 0

2 0 2 0

( ) ( )( ) ( )

a x x b y ydy fdx a x x b x x

1.24

which yields the DE

1 1

2 2

a u b vdy fdx a u b v

after the translation of axes of the form

0 0, .u x x v y y

Homogenous Equation

1.25

Example: Solve

( 3 6) ( 2) 0.x y dx x y dy

Solution: 3 14 0.

1 1

1, 3.u x v y

From 3 6 02 0

x yx y

we obtain 0 01, 3.x y

Hence, we let

1.26

The differential equation for v is

( 3 ) ( ) 0,u v du u v dv

or33 .1

vdv u v u

vdu u vu

The above equation is homogenous, so we let .vz

u

Then 3 ,1

dz zz u

du z

or

23 2 .1

dz z zu

du z

1.27

Separating variables gives

2

1 1 ,2 3

zdz du

z z u

21 ln | 2 3 | ln | | ,2

z z u C

from which it follows that 2 22 3 .z z Cu When we substitute back in for z, u, and v, we find

2 2 2 2( ) 2 3 , 2 3 ,v vcu v uv u C

u u

2 2( 3) 2( 1)( 3) 3( 1)y x y x C

1.28

1.2 Linear First-Order DE

(1) Definition Linear Equation

1 0( ) ( ) ( )dya x a x y h x

dx

A first-order differential equation of the form

is said to be linear equation.When ( ) 0,h x the linear equation is said

to be homogeneous;

(1)

or inhomogenous.

otherwise, it is non-homogenous

1.29

(2) Standard Form

( ) ( ).dyP x y q x

dx

By dividing both sides of (1) by the lead coefficient

1( ),a x we obtain a more useful form, the standform, of a linear equation:

We seek a solution of (2) on an interval I for which

(2)

both function P and f are continuous.

(3) Variation of parameters

1 0( ) ( ) ( )dya x a x y h x

dx

1.30

I. The Homogeneous Equation ( ) 0.dyP x y

dx

This is a separable equation.

(3)

Writing (3) as ( ) 0dyP x dx

y and integrating .

Solving for y gives ( ) .P x dx

y Ce

II. The Nonhomogeneous Equation ( ) ( ).dyP x y q x

dx (2)

Method of variation of constants

1.31

II. The Nonhomogeneous Equation

( ) ( ).dyP x y q x

dx (2)

Method of variation of constantsThe basic idea is the constant C in the general solution of the homogeneous equation (3) is replaced by a function C(x). The calculation of an appropriate choice of C(x) gives a solution of the nonhomogeneous equation (2) .

( )( ) P x dxy C x e

Substituting into (2) gives ( ) ( )( ( ) ( ) ( )) ( ) ( ) ( )P x dx P x dx

C x C x P x e P x C x e q x

1.32

so ( )( ) ( )P x dx

C x e q x

Hence

( )or ( ) ( )P x dxC x e q x

( )( ) ( ) ,

P x dxC x e q x C

Thus if (2) has a solution, it must be of form (4).

and ( ) ( ) ( )

( ) ( )

( )

( ( ) ).

P x dx P x dx P x dx

P x dx P x dx

y Ce e e q x dx

e e q x dx C

(4)

that is a general solution of equation (2).

Conversely, it is easy to verify that (4) constitutes one-parameter family of solutions of equation (2),

1.33

Example Finding the general solution to

The associated homogeneous equation is

1( 1) ( 1) .x ndyx ny e x

dx

Solution:We write the differential equation in standard form

( 1) .1

x ndy ny e x

dx x

0.1

dy ny

dx x

Separating variable, we find

1.34

substituting

0,1

dy ndx

y x

( 1) .ny C x So the general solution to the homogeneous equation is

By the method of variation of parameter,

( )( 1)ny C x x into the equation gives1( ) ( 1) ( 1) ( ) ( )( 1) ( 1) .

1n n n x ndc x n

x n x c x c x x e xdx x

So ( ) ,xdc x edx

1( ) .xc x e c

Thus the general solution to 1( 1) ( 1) .x ndyx ny e x

dx

1.35

is 1( 1) ( ).n xy x e c

1.41

Bernoulli Equations ( ) ndy P x y Q x ydx

Remark: when n=0,1, equation (9) is also a

(9)

linear equation.For 1,n dividing equation (9) by ny

yields 1( ) ( ).n ndyy P x y Q x

dx

Taking 1 ,nv y we find via the chain rule that

(10)

(1 ) ,ndv dyn y

dx dx

1.42

(1 ) ,ndv dyn y

dx dx

and so equation (10) becomes

1 ( ) ( ).1

dvP x v Q x

n dx

Linear

equation

Example : Solve 355 .2

dyy xy

dx

Solution: This is a Bernoulli equation with n=3, 5( ) 5, ( ) .

2x

P x and Q x

(11)

1( ) ( ).n ndyy P x y Q x

dx

1.43

We make the substitution 2 .v y

Since 32 ,dv dyy

dx dx the transformed equation is

1 55 ,2 2

dvv x

dx 10 5 .dv

v xdx

This is a linear equation, its solution is 10 10 10 10

10

( 5 ) ( 5 )

1 .2 20

dx dx x x

x

v e xe C e xe dx C

xCe

1.44

Substituting 2v y give the solution

2 101 .2 20

xxy Ce

0y Not included in the last equation is the solution

that was lost in the process of dividing

(11) by 3 .y

10 10 10 10

10

( 5 ) ( 5 )

1 .2 20

dx dx x x

x

v e xe C e xe dx C

xCe

1.3 Exact Differential Equations 1.3 Exact Differential Equations and Integrating factorsand Integrating factors

1.46

Exact Differential EquationsExact Differential Equations

• Recall that the total or exact differential of a function u(x,y) is:

dyyudx

xudu

0),(),( dyyxNdxyxM

• We will use the concept of exactness to study differential equations of the form:

1.47

1 2( , ) ( ),u x y C R

( , )du x y Mdx Ndy

(exact differential)u udx dy du

x y

• Compare the following

We say an ODE is exact if there exists a function

( , ) and ( , ) .u uM x y N x y

x y

Note this means that we can now write our ODE as:

That is

such that

( , ) ( , ) 0 (ODE)M x y dx N x y dy

1.48

In this case, its solution (in implicit form) is given by

( , )u x y C

NyuandM

xu

Remember we need:

Note this means that we can now write our ODE as:

( , ) 0du x y

1.49

Testing for ExactnessTesting for Exactness

• In practice we often don ’ t know about u, only about M and N and it is hard to check that

NyuandM

xu

• We want a technique of testing for exactness based on knowing M and N, not u

1.50

• Let ’ s check the derivatives of M and N:

yxu

yu

xxNand

xyu

xu

yyM

22

xN

yM

• A necessary and sufficient condition for exactness is:

If M(x,y) and N(x,y) are continuous functions and have continuous first partial derivatives on some simply connected region of xy-plane, then

1.51

Solving Exact EquationsSolving Exact Equations• Once we have checked the equation is exact

it can be readily solved by evaluating either:

( ) from uu Mdx g y M

x

( )=N from u uMdx g y N

y y y

1.52

( )=M from u uNdy h x M

x x x

or ( ) from uu Ndy h x N

y

1.53

Exact Equations: ExampleExact Equations: Example2 2(2 sec ) ( 2 ) 0.xy x dx x y dy

2 2( , ) 2 sec ( , ) 2 .Here M x y xy x and N x y x y

Since 2 ,M Nx

y x

Example Solve Solution

the equation given is exact.

From 22 sec ,uxy x

x

we have that2 2( , ) (2 sec ) ( ) tan ( ).u x y xy x dx g y x y x g y

1.54

Next we take the partial derivative of u with respectto y and substitute 2 2x y for N:

2 2( ) 2 .x g y x y 2Thus ( ) 2 , and ( ) .g y y g y y

Here we drop the constant of integration that technically should be present in g(y) since it will just get absorbed into the constant we pick up in the next step.

2 2( , ) tanu x y x y x y Hence So, the implicit solution to the differential equation is

2 2tan .x y x y C

1.55

Exact Equations: ExampleExact Equations: Example

3 2 2 33 3M x xy and N x y y

6 6 ( )M Nxy and xy exact

y x

3 2 2 3( 3 ) (3 ) 0x xy dx x y y dy

2 3

2 2 4

( ) (3 ) ( )

3 1 ( )2 4

u Ndy h x x y y dy h x

x y y h x

1.56

Exact Equations: ExampleExact Equations: Example• Solve for k(x) and then u:

2 3 23 3u dkM xy x xy

x dx

4

4 2 2 4

41 ( 6 )4

xh c

u x x y y c

1.57

Example Find the solution for the following IVP.

Solution

Let ’ s identify M and N and check that it ’ s exact.

So, it ’ s exact.

3 3 3 2 33 1 (2 3 ) 0.xy xy xyy e ye xy e y

3 3 2 3 3 3

3 2 3 2 3 3 3

3 1, 9 9

2 3 , 9 9

xy xy xyy

xy xy xy xyx

M y e M y e xy e

N ye xy e N y e xy e

(0) 1.y

1.58

With the proper simplification integrating the second one isn ’ t too bad. However, the first is already set up for easy integration so let ’ s do that one.

Differentiate with respect to y and compare to N.

So, we get

Recall that actually h(y) = k, but we drop the k because it will get absorbed in the next step. That gives us h(y) = 0.

3 3( , ) (3 1) ( )xyu x y y e dx h y 2 3 ( )xyy e x h y

3 2 3( , ) 2 3 ( )xy xyyu x y ye xy e h y N

3 2 32 3xy xyN ye xy e ( ) 0 ( ) 0h y h y

1.59

Therefore, we get

The implicit solution is then

Applying the initial condition gives 1 = CThe implicit solution is then

2 3( , ) .xyu x y y e x

2 3 xyy e x C

2 3 1.xyy e x

1.60

Remember Exact EquationsRemember Exact Equations

• Alternate method: we can often rearrange linear first order ODEs into the form:

( , ) ( , )

0

u uM x y dx N x y dy dx dy

x ydu

xN

yM

(Necessary and sufficient)

• We can check whether an equation in this form is exact by checking if

1.61

3 2 2 3( 3 ) (3 ) 0x xy dx x y y dy

Solve the following DE

Solution: Here 3 2 2 33 3M x xy and N x y y

6 6M Nxy and xy

y x

it ’ s exact.

We first check to see if we have an exact equation.

Since

1.62

3 2 2 3( 3 ) (3 )x xy dx x y y dy Note that

So the general solution is

4 2 2 41 ( 6 )4

x x y y C

3 2 2 3(3 3 )x dx xy dx x y dy y dy 3 2 2 2 2 33 3( )

2 2x dx y dx x dy y dy

4 2 2 41 3 1( )4 2 4

d x x y y

1.63

2 21 ( )2

xdx ydy d x y )( 22

22yxd

yx

ydyxdx

(xy)dxy

ydxxdy ln

)(2 xyd

xydxxdy

)](ln[xyd

xyydxxdy

)](tan[ 1

22 xyd

yxydxxdy

( )xdy ydx d xy ( )dx dy d x y

1.66

Example. Solve the homogeneous DE

Solution: This equation can be written in the form

( ) ( ) 0.y x dx x y dy which is an exact equation.In this case, the solution in implicit form is

i.e. ,

2 21 1 .2 2

xy x y C

2 22 .xy x y C

dy x ydx x y

1.67

Integrating FactorsIntegrating Factors• If the equation is not exact we can consider

( , ) ( , ) 0 ( , )

0

M x y dx N x y dyx y

Mdx Ndy

( , ) 0x y

such that the new equation is exact:multiplying it by a function

1.68

Finding Integrating FactorsFinding Integrating Factors• For exactness we require:

( )N MM N

y x x y

( ) ( ) M Ny x

M NM N

y y x x

This equation can be simplified in special cases, two.of which we treat next

1.69

where M Ny x

N

is just a function of x.

Let 1 ( ) M N

xN y x

d M NN

dx y x

( )x • If we choose

M N

d y xdx N

( )N MM N

y x x y

1.70

Now solve forNow solve for

1 ( )d x dx

1 ( )d x dx

1 1 1 ( )d M N dx

dx N y x dx

( )ln ( ) x dxx dx e

1.71

Conversely, ifM Ny x

N

is just a function of x.

Then

Let 1 ( ) M Nx

N y x

( )x dxe

is an integrating factor for Equation.

1.72

In a similar fashion, if equation has an integrating

M Ny x

M

is just a function of y.

Conversely, ifM Ny x

M

is just a function of y.

Let ( ) .

M Ny xy

M

( )y dye

is an integrating factor for Equation.

Then

factor that depends only y, then

1.73

Integrating Factors: ExampleIntegrating Factors: Example

2 22 , ,M x y N x y x

1 2 1 .M Nxy

y x

2 2(2 ) ( ) 0x y dx x y x dy Example: Solve

Solution:

The equation is not exact. We compute

2

1 (2 1) 2(1 ) 2 .(1 )

M Nxy xyy x

N x y x x xy x

1.74

So an integrating factor for the equation is given by 2

2 .dx

xe x

When we multiplying by2 ,x

we get the exact equation2 1(2 ) ( ) 0.yx dx y x dy

Solving this equation, we obtain the implicit

solution 212 .

2y

x yx C

1.75

Hence, 2

122y

x yx C and 0x are solutions to the given equation.

Note that the solution x=0 is lost in multiplying by

2 .x

1.76

Linear Differential EquationsLinear Differential Equations

• Recall a differential equation is linear if it can be written:

• If q(x)=0 the equation is homogeneous, otherwise the equation is nonhomogeneous.

' ( ) ( )y p x y q x

1.77

Homogeneous Linear CaseHomogeneous Linear Case

• Separating variables and integrating:

dxxp

cexy

cdxxpy

dxxpy

dyyxpy

)()(

~)(||ln

)(

0)('

1.78

Nonhomogeneous Linear CaseNonhomogeneous Linear Case

• Nonhomogeneous linear equations can be solved using the integrating factor method. First rewrite the differential equation as:

( ) 0py q dx dy , 1M py q N i.e.

1.79

Finding the integrating factorFinding the integrating factor• To find the integrating factor we first compute

1 ( ) (1)1

py qy x

1 ( , 1)M NM py q N

N y x

, where is a function of p p x

1.80

Nonhomogeneous Linear CaseNonhomogeneous Linear Case

• We can now find an integrating factor

( ) pdxx e

'

( ' )pdx pdx pdxe y py e y e q

• Now multiply the differential equation by the integrating factor:

1.81

Nonhomogeneous Linear Nonhomogeneous Linear CaseCase

pdx pdxe y e q dx c

( )pdx pdxy e e qdx c

( ) (Integrate both sides w.r.t. )pdx pdxe y e q x

1or ,

where ( ) .

( )( ( ) )pdx

y x x qdx c

x e

1.82

Integrating Factors: ExampleIntegrating Factors: Example

, 2 ,yM y N x ye

1 2 .M Ny x

(2 ) 0yydx x ye dy Example: Solve

Solution:

The equation is not exact. We compute

1 2 1 .

M Ny x

M y y

Here

1.83

So an integrating factor for is given by 1

.dy

ye y

When we multiplying by ,y we get the exact equation

Solving this equation, we obtain

2 2(2 ) 0yy dx xy y e dy

as the solution in implicit form.

2 2( 2 2) yxy y y e C

1.84

In general, integrating factors are difficult to uncover. If a differential equation does not have one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution are recommended.

1.85

Example : Solve 21 ( ) ( 0).dy x x

ydx y y

This differential equation is not exact, and no integrating factor is immediately apparent.

Note however, that if terms are strategically regrouped, the DE can be rewritten as

Solution: Rewriting this equation in differential form, we have 2 2( ) 0.x x y dx ydy

2 2( ) 0.xdx ydy x y dx

1.86

2 2( ) 0.xdx ydy x y dx

The first group of terms has many integrating factors (see Table 2). One of these factors, namely

2 2

1( , ) ,x yx y

is an integrating factor for the entire equation.

Multiplying (1) by

(1)

2 2

1( , ) ,x yx y

we find2 2

( ) 0.xdx ydydx

x y

(2)

1.87

Since (2) is exact, it can be solved using the steps described previously.

Alternatively, we note from Table 1,

2 2

2 2

( )xdx ydyd x y

x y

so that (2) can be rewrite as

2 2 0.d x y dx

Integrating both sides of this last equation, we find

1.88

2 2 .x y x c

or equivalently,

2 ( 2 ).y c c x

1.89

Solve ( ) 0.ydx y x dy

Solution: Here ,M y N y x

and, since 1, 1,M Ny x

The differential equation is not exact.

Since 2M Ny x

M y

is a function of y alone.

we have an integrating factor

1.90

2

2

1( , ) .yx y ey

Multiplying the given DE by 2

1( , ) ,x yy

we obtain the exact equation

2

1 ( ) 0,y xdx dy

y y

or equivalently, 2

1 0,ydx xdydy

y y

1.91

Integrating both sides of this last equation, we find

ln | | .xy c

y

(ii) Note that the differential equation can be rewritten as

( ) 0.ydx xdy ydy

The first group of terms has many integrating factors (see Table 1). One of these factors, namely

1.92

2

1( , ) ,x yy

is an integrating factor for the entire equation.

Multiplying (1) by

we find2

( ) 1 0.ydx xdydy

y y

2

1( , ) ,x yy

Integrating both sides of this last equation, we find

ln | | .xy c

y

1.93

Now let yv

x and recall that .dy dv

v xdx dx

With these substitution, equation (3) becomes

2

11

dv v v dxx v or x dv

dx v v x

(iii) Rewriting this equation in the derivative form,

.dy ydx x y

then we see that the equation (3) is homogeneous.

(3)

1.94

The above equation is separable, and , by separating the

variables and integrating, we obtain

1 ln | | ln | | .v x Cv

,

ln | | 0

yFinally we substitute for v to get

xx

y Cy

1.95

(iv) We rewrite the differential equation in the form

1,dx x y dx xor

dy y dy y

which is linear equation.Its solution is

1 1

( ) (ln | | ).dy dy

y yx e e dy c y y c

Also note that x=0 is a solution.

1.96

Table 1

1.97

Table 2

1.98

1.99

Exercise Solve 2( ) (1 ) 0.x y dx y x dy

1.100

1.101

1.102

1.103

1.104